Question

Show that the equation of the line passing through (acos³theta, asin³theta) and perpendicular to the line xsectheta + ycosectheta = a is xcostheta - ysintheta = acos2theta (not cos square)

Answer 1

Dear Student,

Let the equation of line be y = mx + c

Given line: x secθ + y cosecθ = a

Slope of the line = $-\frac{sec\theta }{\cos ec\theta }$
Slope of perpendicular line (m) = $\frac{-1}{-{\displaystyle \frac{sec\theta }{\cos ec\theta }}}$
m = $\frac{\cos ec\theta }{sec\theta }$ = $\frac{\cos \theta }{\sin \theta }$
Now, the line passes through (a cos^​3θ , a sin³θ)

a sin³θ = $\frac{\cos \theta }{\sin \theta }$ (a cos³θ) + c
c = a sin³θ - $\frac{a{\cos }^4\theta }{\sin \theta }$
c = $\frac{a{\sin }^4\theta -a{\cos }^4\theta }{\sin \theta }$
c = $\frac{a\left({\sin }^2\theta -{\cos }^2\theta \right)}{\sin \theta }$
c = $-\frac{a\cos 2\theta }{\sin \theta }$
Therefore, equation of line is : y = $\frac{\cos \theta }{\sin \theta }$ x $-\frac{a\cos 2\theta }{\sin \theta }$
y sinθ = x cosθ - a cos2θ

Regards